NEET 2024 Physics Question 43 Solution
Question
A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14):
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A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14):
Peak current I₀ = V₀ ω C = (√2 × 210) × (2 × 3.14 × 50) × 10⁻⁵ = 297 × 314 × 10⁻⁵ ≈ 0.93 A!
The voltage rating given for an AC power supply (210 V) represents its root-mean-square (RMS) voltage. The peak voltage V₀ is related to V_rms by:
For source frequency f = 50 Hz with π = 3.14:
The opposition offered by the capacitor to alternating current is the capacitive reactance:
The peak current I₀ in the pure capacitive circuit is:
Multiplying the values:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Overestimates current.
Severe calculation error.
Represents the RMS current I_rms (I_rms = 210 × 314 × 10⁻⁵ = 0.659 A, or with arithmetic error 0.58 A).
Exact evaluation of peak current I₀ = √2 V_rms ω C.
Over 40% of students calculate I = V_rms / X_C = 210 × 314 × 10⁻⁵ ≈ 0.66 A, and hunt for an option near it, missing the keyword "PEAK current".
"In a capacitive AC circuit, peak current is I₀ = V₀ / X_C = V₀ ω C, where V₀ = √2 V_rms."
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